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lyudmila [28]
3 years ago
11

An npn BJT has emitter, base, and collector doping levels of 1019 cm????3, 5 1018 cm????3, and 1017 cm????3, respectively. It is

biased in the normal active mode, with an emitter-base voltage of 1V. If the neutral base width is 100 nm, the emitter is 200 nm wide, and we have negligible base recombination, calculate the emitter current, emitter injection efficiency, and base transport factor

Engineering
1 answer:
Darina [25.2K]3 years ago
8 0

Answer:

Explanation:

The answer to the given problem is been solved in the fine attached below.

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By balancing information security and access, a completely secure information system can be created.A. TrueB. False
Inessa05 [86]

Answer: true

Explanation:

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What is the process of sharing carefully planned information to an audience?
maksim [4K]
Presentation, or speech I’d imagine
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External crack of length of 3.0 mm was detected on the surface of the shaft of wind turbine made from 4340 steel. The diameter o
sveticcg [70]

Answer:

The correct answer is "K_c=6.0369 \ MPa\sqrt{m}".

Explanation:

Given:

Maximum load,

P = 50,000 N

Crack length,

a = 3mm

or,

  = 3×10⁻³ m

Diameter,

d = 32 mm

As we know,

⇒  Maximum stress, \sigma=\frac{P}{A}

                                      =\frac{50000}{(\frac{\pi}{4}\times 32^2)}

                                      =62.20 \ N/mm^2

Now,

⇒  Fracture tougness, K_c=Y \sigma\sqrt{\pi a}

On substituting the values, we get

                                           =1\times 62.20\times \sqrt{3.14\times 3\times 10^{-3}}

                                           =6.0369 \ MPa\sqrt{m}

4 0
2 years ago
1- The preexponential and activation energy for the diffusion of iron in cobalt are 1.1×10-5 m 2 /s and 253,300 J/mol, respectiv
n200080 [17]

The temperature at which the diffusion coefficient have a value of 2.1×10-5 m 2 /s is  -47078 K.

Using the relation;

logD = logDo - Ea/2.303RT

D = diffusion coefficient

Do =  preexponential

Ea = activation energy

R = gas constant

T = temperature

Substituting values;

log(2.1×10-5)= log (1.1×10-5 ) - 253,300/2.303 × 8.314 × T

log(2.1×10-5) -  log (1.1×10-5 ) =  - 253,300/2.303 × 8.314 × T

log[2.1×10-5/1.1×10-5] = - 253,300/2.303 × 8.314 × T

0.281 × (2.303 × 8.314 × T) = - 253,300

T =  - 253,300/2.303 × 0.281 × 8.314

T = -47078 K

Learn more: brainly.com/question/14283892

4 0
2 years ago
The Resistance of wire of length
mojhsa [17]
Resisitivty = RA/L = 7*10^-8 ohm meters
Conductivity is the inverse= 14.3*10^6 S/m
4 0
2 years ago
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